CHEMICAL EQUILIBRIUM

 

Hints for Subjective Problems

 

LEVEL – I

1. Equate the concentration of A and D at equilibrium.
2. We can start the reaction taking initial equilibrium of CO as 1.
3. Convert MPa into atmosphere.
4. KMnO4 reacts with Fe+2 at equilibrium.
5. Calculate the partial pressure first.
6. At the same temperature KP or KC remains the same.
7. At the same temperature KP and KC remains the same.

LEVEL – II

1. No. of moles of NO: Total no. of moles at  equilibrium = 1.8 : 100
2. First calculate KC.
3. No. of moles of NaOH = No. of moles of acid at equilibrium.
4. Without dissociation the pressure is first because of the no. of moles of SO2Cl2.
5. Calculate the no. of moles of NH4HS first.
6. a) Find first.

LEVEL – III

1. Compound without lowest vapour pressure is most effective drying agent.
2. With the dissociation of HI, the dissociation of NH4I will also increase.
3. Calculate Q first
4. Pressure over excess solid is due to the gases.
5. Since I2 remains constant so α can be calculated to solve for KC.
6. Pressure of NH3 in both the cases should be same 
7. P ∝ n
8. N2 gas will react with the I2 at equilibrium.
9. Na2S2O3 will react with the I2 at equilibrium.
10. Calculate the KP at 100°C.
11. P ∝ n

 

Solution to Subjective Problems 

 

LEVEL – I

1. Given

A(g) + 2B(g) 2C(g) + D(g)

a a 0 0

(a – x) 2x x

Given that  a – x = x

or (a = 2x)

KC = = 4

∴ KC = 4

2. SO2(g) + NO2(g) SO3(g) + NO(g) KC =16

Initial (mole) 1 1 1 1

At eq. (mole) 1–α 1–α 1+α 1+α

KC = = 16

= 4

α =

∴[NO] = 1+ = = 1.6 M

[NO2] = 1– = = 0.4 M

3. Let the initial moles of CO is 1 and x is the mole of CO reacted before equilibrium

FeO(s) + CO(g) Fe(s) + CO2(g)

              1                           0

              1 – x                     x

Then given that = 0.403 ⇒ x = 0.287

Mole fraction of remaining CO at equilibrium = = 1 – 0.287 = 0.713

4. C2H4 + H2O C2H5OH

1 10 0 at initial

1 – x 10 – x x at equilibrium

Total moles at equilibrium = 11 – x

= K

= 0.306

P =- ⇒ x = 0.732

∴ % dissociation = 73.2%

5. Initial conc. of AgNO3 = = 0.075

Initial conc. of  Fe2+ solution = = 0.545

Final conc. of Fe2+ solution = = 0.499

Ag+(g) + Fe2+(aq)  Fe3+(aq) + Ag(s) 

0.029 0.499 0.046

KC = = = 3.178

6. In the given reaction 87%  of the acid is  consumed at equilibrium

C6H5COOH(l) + C2H5OH (l) C6H5COOC2H5(l) + H2O(l)

Initially 1 3 0 0

At equilibrium 1–0.87 3–0.87 0.87 0.87

∴ KC = =  = 2.73

Let n be the number of moles of acid consumed when 1 mole of the acid is mixed with 4 moles of ethanol.

C6H5COOH (l) + C2H5OH (l) C6H5COOC2H5(l) + H2O(l)

At equilibrium 1–n 4–n n     n

KC = = 2.73

= 2.73

On solving, we get n = 0.90 or 6.99

6.99 is not possible as we started with 1 mole of the acid 

∴% of acid consumed  = 90%

7. N2O4 2NO2

Initial (mole) 100 0

At eq. (mole) (100–25) 50 = 75

At eq. (p.p.) 1× = 1× =

∴ KP = = 0.267 atm.

(ii) Suppose x % N2O4 dissociates at 0.1 atm pressure 

N2O4  2NO2

Initial (mole ) 100 0

At eq. (mole) 100–x 2x   total moles at equilibrium = 100 + x 

At eq. (p.p.)    

∴KP =

On solving, we get  x = 63.25 %

8. Let 1 mole of PCl5 be taken initially. If ‘x’ moles of PCl5 dissociate at equilibrium, its degree of dissociation = x

Moles	PCl5	PCl3	Cl2
Initially	1	0	0
At equilibrium	1-x	x	X
Total moles	1 – x + x + x = 1 + x

P = 5 atm Kp = 0.82

=  

Kp =

Kp =

x = 0.375 (0r 37.5%)

Now the new pressure P = 10 atm

Let y be the new degree of dissociation. As the temperature is same (250°C), the value of Kp will remain same. i.e. the same manner. Proceeding in the same manner.

Kp =

⇒ y = or y = 0.275 (or 27.5%)

Note 1: by increasing pressure, degree of dissociation has decreased, i.e., the system shifts to reverse direction. Compare the result by applying Le Chatelier’s principle.

9. a) We know that

⇒ k350= 4.3

2. b) Equilibrium constants at different temperature and heat of the reaction are related by the equation
   =
   303 log = =

log

10. a) First determine the equilibrium constant

KC for H2 + I2 2HI

KC =

When 0.3 mol of HI are added, equilibrium is disturbed. At the instant,

[HI] = 0.49 + 0.3 = 0.79 M

Q > KC, since KC =

⇒ backward reaction dominate and the equilibrium shifts to the left.

Let 2x = concentration of HI consumed (while going left) then concentration of each of H2 and I2 formed = x

[HI] = 0.79 – 2x, [H2] = 0.08 + x, [I2] = 0.06 +x and KC = 50

⇒ KC =

⇒ 46x2 + 10.2x – 0.35 = 0

x = 0.033 or – 0.25 (neglecting the –ve value)

Finally, the equilibrium concentrations are:

[HI] = 0.79 – 2x =0.79 – 0.033 × 2 = 0.724 M

[H2] = 0.08 + x = 0.08 + 0.033 = 0.113 M

[I2] = 0.06 + x = 0.06 + 0.033 = 0.093

7. b) i) Total no. of moles at equilibrium = 7.6 × 10–3 + 6.6 × 10–3 = 1.42 × 10–2 moles

If 1.42 × 10–2 moles of argon have been added and the volume has been doubled there will not be any change in no. of moles/litre at equilibrium. Hence no effect

1. ii) At constant pressure the equilibrium will shift in the direction where no. of moles are decreasing hence to the left.

iii) No effect at constant volume

 

LEVEL – II

 

1. N2 + O2 2NO

79 21 0 at equilibrium

(79 – x) (21 – x) 2x at equilibrium

Given that

x = 0.9

KC = = 2.1 × 10–3 = KP

2. Kp = KC(RT)Δn

KC = = 1.7 × 10–4

F2 2F

1 0 at initial

(1 – x) 2x at equilibrium

1.7 × 10–4 = x << 1

x2 = 0.425 × 10–4

x = 0.65 × 10–2 = 0.65

3. C2H5COOH(l) + C2H5OH(l) C2H5COOC2H5(l) + H2O(l)

0.5 0.5 0 0 at initial

(0.5 – x) (0.5 – x) x x at equilibrium

= 7.36 ⇒ x = 0.36

Moles of ethyl propionate = 0.36

Weight of ethyl propionate = 0.36  × 102 = 36.72g

4. C2H5OH9l) + C3H7COOH(l) C3H7COOC2H5(l) + H2O(l)

at initial 0.1 0.1 0 0

0.1 – x 0.1 – x x x

Given that

0.1 – x = 0.85 × 100 × 10–3

0.1 – x = 0.085 ⇒ x = 0.75

K = = 9

5. 2AB2(g)   2AB (g) + B2(g) 

Initial (mole) 1 0       0

At eq. (mole) 1–α α       α/2   Total moles at equilibrium = (1+α/2)

At eq. (p.p)  

KP =

KP =

But 1>> α

∴ KP =

α =

6. 2H2S(g)    2H2(g) +  S2(g) 

2(1–α) 2α     α

Where α is the degree of dissociation 

Total number of moles at equilibrium = 2–2α + 2α+α = 2+α

= , = , =

KP =

If α = 0.055 and P = 1 atm

KP = = 9.114 × 10–5

KP = KC (RT)Δn, [Here Δn =1]

KC = = 3.71 × 10–6

7. i) If there is no dissociation the pressure can be calculate as PV = nRT

PV = nRT

P × 1 = × 0.0821 × 375 = 1.53 atm

1. ii)  Δn = 1

 KC= ⇒ 2.4  =

⇒ Kp = 73.8

SO2Cl2(g) SO2(g) + Cl2(g)

1.53 0 0 at initial

1.53 – x x x at equilibrium

Kp = = 73.8 ⇒ x = 1.5

∴ Partial pressure of SO2Cl2 = 1.53 – x 

= 1.53 – 1.5 = 0.03 atm

1. c) Total pressure at equilibrium = 1.53 – x + x + x

= 1.53 + x = 1.53 + 0.3 = 1.53 atm

8. NH4HS(s) NH3(g) + H2S(g)

0.018 0.018

KC = (0.18)2 = 3.24 × 10–4

KP = KC(RT)Δn  Δn = 2

= 3.20 × 10–4 × (0.821 × 300)2

= 0.19 atm

1. ii) No effect
2. a) 2NO(g) + Br2(g) 2NOBr(g)

Initial pressure   98.4     41.3 0

At eqlbm. 98.4-x    41.3 – x

The total pressure at equilibrium is 110.5 torr.

      ∴ 98.4-x + 41.3-+ x = 110.5

x = 58.4 torr

Now,  1 atm = 760.4 torr,      ∴ x = 7.68 × 10-2 atm.

pNOBr = 7.68 × 10-2 atm  ;     pNO = 98.4-x = 40 torr = 5.26 × 10-2 atm.

pBr2 = 41.3 – = 12.1 torr = 1.59 × 10-2 atm.

=  

ΔGo = -2.303 RT log K =  -2.303 (1.99) (300) (log 134)  =  -2.92 k cal  = 12.2 k J.

[If R is used as 1.99 cal/mol K, then ΔGo will be in cal. If R is used as 8.314 J/mol K, then ΔGo will be in Joules. But KP must always be in (atm) Δn.]

2. b) B = A, ΔG10 = {(− 2.303× 8.314×448)log(1.3/95.2)}=15.998 kJ

B = C, ΔG20 = {(− 2.303× 8.314×448)log(3.5/95.2)}=12.305 kJ . 

Stability order B > C > A

 

10. a) PCl5(g)  PCl3(g) + Cl2(g)

We are given the vapour densities at equibilibrium at 200oC and 250oC.

The initial vapour density will be the same at both the temperatures as it would be
∴  Initial vapour density =
Vapour density at equilibrium at 200oC = 70.2

∴
=
∴ α = 0.485
At 250oC, 1 + α = ,  ∴ α = 0.8

1. b) Let us assume that we start with C moles of N2O4(g) initially.

N2O4(g)    2NO2(g)

Initially C     0

At equilibrium C(1−α)   2Cα

where α is the degree of dissociation of N2O4(g) 

Since,

Initial vapour density = 

Since vapour density and actual density are related by the equation,
V.D. = = =  26.25

∴ 1 + α =

∴  α = 0.752

∴ Kp =

KP=   5.2 atm.

 

LEVEL – III

1. a) i) kP = = 6.89 × 10–12

⇒ = 1.62 × 10–3 atm

5. ii) kp = = 5.25 × 10–13

⇒ = 3.5 × 10–3 atm

iii) kp = = 4.08 × 10–25

⇒ = 3.64 × 10–3 atm

1. b) SnCl2⋅2H2O is most effective drying agent since it was lowest vapour pressure of water in equilibrium with it.
2. c) The hydrate Na2SO4.10H2O will loose water (efflorescence) below 3.04 × 10–3 atm (2.77 torr) and relative humidity can be calculated as

% relative humidity =

= = 60.5%

60. d) The dehydrated salt will absorb water above 60.5% relative humidity.
61. P = = 0.3618 atm

= 0.181 atm

Kp = (0.181)2 = 0.033 atm2

NH4I(s) NH3(g) +   HI(g)

      (0.181 + p)       (0.181  + p – p′) Kp = 3.3 × 10–2 atm2

2HI(g) H2(g) + I2(g)

(0.181 + p – p′) kp = 0.015

0.033 = (0.018 – p) (0.181 + p – p′)

0.015 =

⇒ p′ = 0.04 atm

p = 0.026 atm

∴ total pressure = 2(0.181 + p) = 0.414 = 314.64 mm Hg

3. = – 9.346 + …(1)

Similarly

logKp2 = – 9.346 + …(2)

Subtracting (1) and (2)

=

= 9888

Comparing from

= 9888

ΔH = 9888 × 2.303 × 8.314 = 189244.73J = 189.244 kJ 

4. NH4HS(s) NH3(g) + H2S(g)

Kp = KC(RT)Δn

KC = = = 1.2× 10–4 Δn = 1

Q = = 0.024

 KC < Q backward reaction is taking place 

initial pressure of NH3 = = 3.037 atm

initial pressure of H2S = = 7.29 atm

NH3(g) + H2S(g) NH4HS(s)

3.037 7.29                 0 at initial

(3.037 – x) (7.29 – x) at equilibrium

Kp′= ⇒ x = 3 atm

= 0.0257 atm

= 4.278 atm

5. i) Volume at NTP = ml = 7.799

∴ Mole at NTP =

Let the observed molecular weight of selenium be M

∴ No. of moles =

Hence

M = 217.6

∴ Observed density = = 108.8

Theoretical density = = 237 (molecular weight of Se6 = 79 × 6)

(i.e. where there is no diss)

Since 1 molecule of selenium produces 3 molecules

∴ n = 3

Substituting the above values in equilibrium reaction we get

Degree of dissociation of Se6 = = 0.59

1. ii) Se6(g) 3Se2(g)

1 0 initial mole (suppose)

(1 – x) 3x mole at equilibrium (x = deg. of dissociation)

Total moles at equilibrium = 1 – x + 3x = 1 + 2x

Kp = × P2

Substituting x = 0.59 and atm

Kp = 0.1687

iii) Kp = Kc(RT)Δn

0.1687 = Kc(0.0821 × 973)2 (Δn = 3 – 1 = 2)

KC = = 0.2645 × 10–4

6. X(s) A(g) + C(g) at equilibrium A & C are in equal proportions, so  their pressures will be same.

Also PA + PC = 40 ⇒ PA = PC = 20 mm ⇒ Kp = PA.PC = 202 = 400 mm2

Similarly Y(s) B(g) + C(g)

PB = PC = 30 mm ⇒ Kp = PB . PC = 302 = 900 mm2

1. i) Now for a mixture of X and Y, we will have to consider both the equilibrium simultaneously.

X(s) A(g) + C(g) and

Y (s) B(g) + C(g)

Let PA = a mm, PB = b mm

Note that the pressure of C due to dissociation of X will also be a mm and similarly the pressure of C due to dissociation of Y will also be b mm.

⇒ PC = (a + b) mm

Kp (for X) = PA . PC = a ( a + b) = 400 (I)

Kp (for Y) = PB.PC = b (a + b) = 900 (ii)

From (I) and (ii), we get

as volume and temperature are constant, the mole ratio will be same as the pressure ratio.

1. ii) The total pressure = PT = PA + PB + PC

= a + b + 9a + b) = 2(a + b)

to find a and b, solve the equations

& a (a + b) = 400

⇒ a = 11.1 mm, b = 24.97 mm ⇒ Total pressure = 2 (a + b) = 72.15

7. I2 + I– I3–

Initial (conc.) 0.1 0

At eq. (conc.) (0.1–α) α

But –α =

0.0492 – α = 0.0013

α = 0.0479

At eq. conc. (0.0013) 0.0521 0.0479

∴ KC =

KC = 707.2

8. CO(g)   + 2H2(g)  CH3OH(g)

Initial (mole) 0.15 a 0

At eq. (mole) (0.15–x) (a–2x) x

(0.15–0.08) (a–0.16) 0.08

But, x = 0.08 mole = 0.07

Total number of moles at equilibrium  =  a – 0.01

number of moles at equilibrium =  = = 0.35

∴ a– 0.01 = 0.35; a= 0.36

At equilibrium [CO] =  ,  [H2] = , [CH3OH] =

∴ KC =

KC = = 178.57 M–2

∴KP = 178.57  (0.082 × 750)–2 = 0.047 atm–2

(b) When no reaction takes place, so total moles = 0.51

∴ Pfinal  =  = 12.546 atm

9. LiCl.3NH3 (s) LiCl.NH3(s) + 2NH3(g)

KP = 9 atm, Vol = 5l , Temperature = 40 °C

Number of moles of LiCl.NH3 = 0.1

2NH3(g) + LiCl.NH3(s) LiCl.3NH3(s)

a–2x 0.1–x x

0.1–x = 0 ⇒ x = 0.1

a–2x = a–0.2

KP′ = =

= 5.1395 (a–0.2)

KP′ =

(a–0.2)2 =   ⇒ a = 0.7837

∴ Minimum number of moles of NH3 required = 0.7837

10. The given equilibria is

N2(g) + 3H2(g) 2NH3(g)

At t=0 4 16 0

At equilibrium 4 – α 16 – 3α 2α

Total gaseous moles at equilibrium = 4 – α + 16 – 3α + 2α = 20 – 2α

Since, pressure has fallen to 9/10 of its original value, hence no. of mole will also fall up to the same extent.

∴ (20 – 2α) = ; α = 1

∴[N2] = = 3 mole/litre

[H2] = = 13 mole/litre

[NH3] = = 2 mole/litre

Kc = = = 6.07 × 10–4

11. COCl2 (g) CO(g) + Cl2(g)

Initial moles 1 0 0

Moles at eq. 1–α α α

Total moles  = 1 + α + a ( Let moles of N2 = a)

K-p = —————– (i) 

By gas equation PV = nRT

P × 100 = (1+α + a) × .0821 × 373

= = 0.224

Putting this value in equation  (i) and since α is very small since Kp  is very low 1 – α ≈ 1

Kp = = 6.7 × 10–9

⇒ α = 1.72 × 10–4

12. First find the value of KC for dissociation of HI from its degree of dissociation

2HI H2 + I2 (degree of dissociation is 0.8)

Concentrations	2HI	H2	I2
Initially	1.0	0	0
At new equilibrium	1.0 – 0.8	0.4	0.4

KC =

Now we have to start with 0.135 mol each of H2 and I2 and the following equilibrium will be established.

H2 + I2 HI with KC = 1/4

Concentrations	H2	I2	2HI
Initially	0.135	0.135	0
At new equilibrium	0.135 – x	0.135 – x	2x

⇒ KC = ⇒ x = 0.135/5 = 0.027

now find the moles of I2left unreacted at equilibrium

n(I2) = 0.135 – 0.027 = 0.108

I2 reacts with sodium thiosulphate (Na2S2O3) as follows:

2Na2S2O3 + I2 Na2S4O6 + 2NaI

applying the mole concept, we have,

2 moles of Na2S2O3 ≡ 1 mole of I2

⇒ 0.018 moles of I2 ≡ 2 × 0.108 = 0.216 moles of Na2S2O3 are used up

⇒ moles = MVn (M = molarity, Va = volume in litres)

⇒ 0.216 = 1.5 V

⇒ V = 0.144 lt = 144 ml

13. Firstly calculate KP at 100°C by using equation 

log 

On solving  we get, (KP)100°C = 0.98 

 

C2H2 + D2O C2D2 + H2O

 

Initial mole 1 2 0 0

At eq. mole 1–α 2–α α α

∴ 0.98 =

0.02α2 + 2.94α – 1.96 = 0

On solving we get 

α = 0.75

14. NH2COONH4(s) 2NH3(g) + CO2(g)

Let P = original equilibrium pressure, from the mole ratio of NH3 and CO2 at equilibrium, we have

&

⇒ Kp =

Now NH3 is added such that, = P

Find the pressure of CO2

Kp = ⇒ ⇒

Total new pressure = Pnew =

⇒

⇒ ratio =

15. It can be seen that as SO2 is not present initially, so equilibrium cannot be established in the forward direction. Therefore it is established from reverse direction. Let n be the increase in partial pressure of O2. Then at equilibrium the partial pressures of SO2, O2 at SO3 are (0+2n), (2+n) and (1–2n) in atm respectively.

2SO2(g)    + O2(g) 2SO3(g)

(0+2n) (2+n) (1–2n)

Also, 

As n is small (because equilibrium constant for the reverse reaction is very small i.e., 1/900), it can be neglected in comparison to 2 and also
1–2n can be taken approximately to 1.

∴ 900 =

Solving for n, we get  n = 0.0118

Hence, = 2+n = 2.0118 atm

= 2n = 0.0236 atm

= 1–2n = 1–0.0236 = 0.9764 atm

 

Solution to Objective Problems 

 

LEVEL – I

1. An increase in temperature favours endothermic reaction i.e. backward reaction if

KOH (s) + aqueous KOH(aq) ΔH = – x

∴ (B)

2. gas liquid ΔH = (–)ve

∴ (A)

3. According to Lechatlier principle.

∴ (C)

4. k for overall reaction = k1 × k2 = 102 × 102 = 1

∴ (C)

5. Kp = KC(RT)Δn

Δn = –1 here ⇒ Kp = 25 × (0.0821 × 523)–1 = 0.6 atm–1

∴ (A)

6. Solid liquid; ΔH = positive 

Increase in temperature favours forward reaction

∴ (A)

7. QK =

An increase in [H+] will show a decrease in [HS–] to maintain constant K.

∴ (D)

8. Since ΔH is positive the reaction is endothermic . As per LeChatlier principle, when  the temperature is raised, it should proceed towards that direction where heat is absorbed i.e.  towards the product side. 

∴(A) 

9. For the reaction H2 + I2 2HI (g) since Δn = 0. Equilibrium constant KC will be uneffected by change in pressure. 

∴ (A)

10. C2H5OH + CH3COOH CH3COOCH3 + H2O

Initially 1 1 0 0

At equilibrium

K = =

∴ (A)

11. Since decomposition of PCl5 is carried out in a closed container. When helium gas is introduced the volume remains constant and there is no change in the partial pressures of PCl5, PCl5 and Cl2. Therefore, nothing happens on introduction of helium gas and constant volume.

∴(D)

12. KP = KC(RT)Δn

KP <KC if Δn<0

For reaction (C) the value of Δn is –2 and for reaction (D) the value of
Δn is–1

∴ (C) & (D)

13. 2AB4(g) A2(g) + 4B2(g) ΔH = –ve

It is an Exothermic reaction and hence  favoured at low temperature. Δn for the reaction is +3. Therefore low pressure  will favour the forward reaction

∴(C)

14. KC = = 7.5 × 102

∴(B)

15. For the backward reaction ΔH = + 40Kcal mol1 for endothermic reactions K increase with increase of temperature

∴ (C)

 

LEVEL – II

1. Volume of ice > volume of water and thus increase in pressure favours forward reaction showing decrease in volume.

∴ (B)

2. Cdiamond Cgraphite

For a given mass Vdiamond <Vgraphite

∴ Thus increase in pressure will favour backward reaction.

∴ (A)

3. N2 + 3H2 2NH3

a a 0 at initial

(a – x) (a – 3x) 2x at equilibrium

⇒ (a – x) > (a – 3x)

∴ (B)

4. N2O4(g) 2NO2(g)

1 ≈ 92g          0 at initial

92 – 92 × 0.2 = 73.6g

= 0.8 mole     0.4 mole at equilibrium

Total moles = 0.8 + 0.4 = 1.2

At constant volume =

⇒ P2 = 2.4 atm

∴ (B)

5. SO2Cl2 SO2 + Cl2

1 0 0 at initial

1 – α α α at equilibrium

, ,

= α2

α = = 0.17

α = 17%

∴ (C)

6. CO2(g) + C(s) 2CO(g)

0.5 0 at initial

(0.5 – x) 2x at equilibrium

∴ 0.5 – x + 2x = 0.8 ⇒ x = 0.3 atm

= 0.5 – 0.3 = 0.2 atm

= 2x = 0.6 atm

kp = = = 1.8 atm

∴ (A)

7. PCl5 PCl3 + Cl2

1 0 0 at initial

1 – α α α at equilibrium

⇒ α = 0.68 ⇒ 68%

∴ (B)

8. For the reaction SO2(g) + O2(g) SO3(g) 

Given KP = 1.7 × 1012 at 20°C and 1 atm pressure 

KP = Kc (RT)Δn; Δn = 1–

1.7 × 1012 = Kc (0.0821 × 293)–1/2

Kc =

Kc = 7 × 1012 × (0.0821 × 293)+1/2 = 8.338 × 1012

∴(C) 

9. For 2A(g) 2B(g) + C(g)  

KP = 1.8 at 700°C

∴ For 2B(g) + C(g) 2A(g) at 700°C

KP = Δn = 2 – (2+1) = –1

KP = Kc(RT)Δn

Kc =

Kc = KP  × (0.0821 × 973)1  = × 0.0821 × 973

Kc = 44.3796

∴ (C)

10. N2O4(g) 2NO2(g) 

t = 0 1 0

t = equilibrium 1–α 2α

∑n = 1 + α

Kp =

Kp =

∴(C)

11. For H2(g) + O2(g) 2SO3(g)

t = 0 2 moles 1 moles 0

t = equilibrium 2(1 – 0.5) 1(1– 0.5) 1 mole

2–1 + 0.5 +1  = 2.5

P′ = P × mF

= = 1 × = 0.20 atm

∴ (D)

12. Enthalpy changed of a reaction is given by 

ΔH = (Ea)f – (Ea)b

Where (Ea)f and (Ea)b are energies of activation for the forward and backward reactions.

ΔH = 12–20 = –8 kJ / mol

KP for the reaction at 25°C = 10 atm. Since KP is expressed in
atmosphere, Δn = +1

QKP = KC (RT)Δn,    KC = = 0.4 M

KC at 40°C is given by 

log = = –0.06719

(KC)40/(KC)25 = 0.85

(KC)40 = 0.85 ×0.4 = 0.34 M

∴(C)

13. 2HI(g) H2(g) + I2(g)

1–α Where α is the degree  of dissociations

α = 2–2α

2+5 α = 2

α = = 0.22

∴(D)

14. N2O4(g) 2NO2(g)

Equilibrium 1–α 2α

Where α is the degree of dissociation 

= ( Q V is constant )

∴ P2 = 2 atm

After dissociation of N2O4 at 600K,

= 2 (1–α) = 2–2α

= 2 × 2α = 4 α

Total pressure = 2–2α+4α = 2+2α

2+2α = 2.4 (Given)

α = 0.2

∴ Percentage dissociation  = 20%

∴(B)

15. N2 + 3H2 2NH3

t = 0 0.2 0.6 0

at = t 0.2–n 0.6–3x 2x

40% of N2 = 0.2 ×0.4 = 0.08

40% of H2= 0.6 ×0.4 = 0.24

∴ number of moles of N2 remaining = 0.2 – 0.08 = 0.12

number of moles of H2 remaining  = 0.6–0.24 = 0.36

number of moles of NH3 formed = 0.16

Total number of moles = 0.12 + 0.36 + 0.16 = 0.64

∴ =

∴(A)

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